Factor and solve \(2x^{2}-8x=0\). Extract the complete GCF, solve every zero-product equation, and substitute each result into the original equation as a check.

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2x² − 8x = 2x(x − 4), so the zeros are x = 0 and x = −4. The second factor gives x = 4, not −4.

2x² − 8x = 2(x − 4)², so x = 4 is a repeated zero. Expanding gives 2x² − 16x + 32, not the original equation.

Factoring out 2x gives 2x(x − 4) = 0. Therefore x = 0 or x = 4; substituting either value into 2x² − 8x gives 0.

2x² − 8x = 2x(x − 4), so the zeros are x = 0 and x = 4. No check of the factorization or solutions is supplied.

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